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A characterization of right triangles II

Recently the following problem was posted on a Facebook group:

Right now we’ll use this problem to obtain another characterization of right triangles, in addition to the ones here.

Stretched image

To start with, here’s an enlarged version of the above image:

In any triangle, the ratio of the length of an internal altitude to the “length” of a right bisector is always sandwiched between 1 and 2.

Strict inequalities

In general, the ratio under consideration lies strictly between 1 and 2.

In triangle ABC, let F be the foot of the altitude from vertex A and M the midpoint of BC. Suppose that the right bisector of BC intersects side AC at N. If AC> AB, PROVE that: \frac{AF}{MN}=\frac{a^2+b^2-c^2}{a^2}.

Details here.

In triangle ABC, let F be the foot of the altitude from vertex A and M the midpoint of BC. Suppose that the right bisector of BC intersects side AC at N. If AC> AB, PROVE that: \frac{AF}{MN}>1.

Details here.

In triangle ABC, let F be the foot of the altitude from vertex A and M the midpoint of BC. Suppose that the right bisector of BC intersects side AC at N. If AC> AB, PROVE that: \frac{AF}{MN}<2.

Details here.

Together we have: 1<\frac{AF}{MN}<2.

Special instance

When the ratio under consideration equals 2.

Suppose that ABC is a right triangle in which \angle B=90^{\circ} and that M is the midpoint of BC. If the right bisector of BC intersects side AC at N and AC> AB, PROVE that: \frac{AB}{MN}=2.

Use similar triangles. Or else use the fact that: \frac{AF}{NM}=\frac{a^2+b^2-c^2}{a^2} from example 1. Take F=B so that b^2=a^2+c^2. Simplify to get \frac{AB}{MN}=2.

In triangle ABC, let F be the foot of the altitude from vertex A and M the midpoint of BC. Suppose that the right bisector of BC intersects side AC at N. If AC> AB and \frac{AF}{MN}=2, PROVE that ABC is a right triangle with \angle B=90^{\circ}.

Details here.

Takeaway

Let ABC be a triangle in which F is the foot of the internal altitude from vertex A, M is the midpoint of side BC, and the right bisector of BC intersects AC (or AB, depending on which is greater) at N. Then the following statements are equivalent:

  1. ABC is a right triangle in which B=90^{\circ}
  2. \frac{AF}{NM}=2.

Tasks

  • (Silver ratio) Suppose that ABC is a triangle containing an obtuse angle B. Let F be the foot of the altitude from C and let the right bisector of AB intersect AC at N.
    1. PROVE that \frac{CF}{NM}> 2, where M is the midpoint of AB.
    2. Find a triangle for which \frac{CF}{NM}=1+\sqrt{2}, the silver ratio.